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# Proofs

asymptotically follows a χ21 distribution. In our case, we have Tn = 0.322 (p-value=0.57).

Therefore, we cannot reject the hypothesisH:ζ12. Testing the hypothesisH :σ12 in model M5 is also a regular problem and therefore the likelihood ratio statistic

Tn= 2

a12,a21max12,φ,σ12

˜ln(a12, a21, ζ1, ζ2, φ, σ1, σ2)− max

a12,a2112,φ,σ

˜ln(a12, a21, ζ1, ζ2, φ, σ, σ) asymptotically follows a χ21 distribution. In our case, we have Tn = 112.8643 (p-value

<0.001). Therefore, we clearly reject the hypothesis H :σ12. These results motivate testing for homogeneity in the model

Xt =ζ+φXt−1Stt, t

iid∼ N(0,1), (4.5.2) where the intercept ζ is treated as a structural parameter, i.e. it is equal in every state, using the MQLRT introduced in Chapter 2. In this case, we can reject the hypothesis of one regime by a p-value < 0.001 (Mn = 54.41). The modified MLE (αb,ζb,φb,bσ1,σb2) is given by (0.248,1.234,0.042,4.943,10.354). The full model conditional MLE ωˆ in model (4.5.2) yields ωb = (ba12,ba21,ζ,b φ,b bσ1,bσ2) = (0.016,0.053,1.210,0.081,5.164,10.371). There-fore, model (4.5.1) as well as (4.5.2) seem to be appropriate models for the series of log returns for IBM stock.

4.6 Proofs 107 extension to autoregressive models of higher order, i.e.p > 1, will be analogous to the case of the variance being a structural parameter instead of a switching parameter. Without loss of generality we assume ζ0 = 0 and σ0 = 1.

Note that we can assume that(ζ1, ζ2, φ, σ1, σ2)are in a small neighorhood of(0,0, φ0,1,1) by Lemma 4.1 wheneverδ0 < α <1−δ0 for any δ0 ∈(0,0.5]. As in Chapter 3 we define

Hn(α) = n−

n

X

t=1

wt

!

log(1−α) +

n

X

t=1

wtlog(α) +p(α)

=: Rn(α) +p(α), where

wt = αg(Xt|Xt−12, φ, σ2)

(1−α)g(Xt|Xt−11, φ, σ1) +αg(Xt|Xt−12, φ, σ2).

Let α = arg maxα∈[0,1]Hn(α). The following lemma shows that the EM-iteration changes the fitted value ofα by no more thanop(1).

Lemma 4.2. Under the conditions of Lemma 4.1 and ifα−αj =op(1)for someαj ∈(0,1), then we have under the null model

α−αj =oP(1).

The proof of Lemma 4.2 is essentially the same as the proof of Lemma 3.4 and is therefore omitted.

In a first step give a stochastic upper bound for

2{pln(α, ζ1, ζ2, φ, σ1, σ2)−pln(0.5,ζb0,ζb0,φb0,σb0,σb0)}. (4.6.1) Note that

2{pln(α, ζ1, ζ2, φ, σ1, σ2)−pln(0.5,ζb0,ζb0,φb0,σb0,bσ0)}

= 2{ln(α, ζ1, ζ2, φ, σ1, σ2)−ln(0.5,ζb0,ζb0,φb0,bσ0,bσ0)}

+ 2{p(α)−p(0.5)}

≤r1n(α, ζ1, ζ2, φ, σ1, σ2) +r2n,

(4.6.2)

where

r1n =r1n(α, ζ1, ζ2, φ, σ1, σ2) = 2{ln(α, ζ1, ζ2, φ, σ1, σ2)−ln(0.5,0,0, φ0,1,1)}, r2n = 2{ln(0.5,0,0, φ0,1,1)−ln(0.5,ζb0,ζb0,φb0,bσ0,bσ0)}.

The last inequality in (4.6.2) follows from the properties of the penalty function p(α).

Letting p(α) ≡ 0, e.g. in the case of a test based on fixed proportions the inequality in (4.6.2) becomes an equality.

Letr1n(α, ζ1, ζ2, φ, σ1, σ2) = 2Pn

t=1log(1 +δt) with δt= (1−α)

g(Xt|Xt−11, φ, σ1) g(Xt|Xt−1; 0, φ0,1) −1

g(Xt|Xt−12, φ, σ2) g(Xt|Xt−1; 0, φ0,1) −1

. Using the inequality log(1 +x)≤x−(1/2)x2+ (1/3)x3 leads to

r1n≤2

n

X

t=1

δt

n

X

t=1

δ2t + (2/3)

n

X

t=1

δ3t. (4.6.3)

For0≤l, s, i≤4we define

ml,s,i = (1−α)ζl121−1)s(φ−φ0)i+αζl222−1)s(φ−φ0)i. Denoting

ζlσs2φig(Xt|Xt−1; 0, φ0,1) = ∂l+s+ig(Xt|Xt−1;ζ, φ, σ)

lζ∂s2)∂iφ

(ζ,φ,σ)=(0,φ0,1)

and expandingg(Xt|Xt−1h, φ, σh), h= 1,2, up to order4, we get δt=

4

X

l+s+i=1

1 l!s!i!

ζlσs2φig(Xt|Xt−1; 0, φ0,1)

g(Xt|Xt−1; 0, φ0,1) +(1)tn (4.6.4) with remainder (1)tn for which

(1)n :=

n

X

t=1

(1)tn =OP(n1/2)





2

X

h=1

X

l+s+i=5 l,s,i≥0

h|l2h−1|s|φ−φ0|i





. (4.6.5)

This is due to the CLT for stationary and ergodic martingale differences.

We reexamine the remainder term (1)n in order to simplify it. To this end we distinguish three cases:

Leti= 0. In this case we havel+s= 5. Following the assessment in Chen and Li (2009), we have by Lemma 4.1 forl = 0,1,2

OP(n1/2) ( 2

X

h=1

h|l2h−1|5−l )

=OP(n1/2) ( 2

X

h=1

2h−1|3 )

4.6 Proofs 109 and for l = 3,4

OP(n1/2) ( 2

X

h=1

h|l2h−1|5−l )

=OP(n1/2) ( 2

X

h=1

h|32h−1|

) . Fori= 1, we have l+s= 4. For l = 0,1, we obtain

OP(n1/2) ( 2

X

h=1

h|l2h−1|4−l|φ−φ0| )

=OP(n1/2) ( 2

X

h=1

2h−1|3 )

, for l= 2

OP(n1/2) ( 2

X

h=1

ζ2h2h−1)2|φ−φ0| )

=OP(n1/2) ( 2

X

h=1

h|(σ2h−1)2 )

,

for l= 3

OP(n1/2) ( 2

X

h=1

h|32h −1||φ−φ0| )

=OP(n1/2) ( 2

X

h=1

{|ζh|5+ (φ−φ0)2} )

which is due to

h|32h−1||φ−φ0| =|ζh2h2h−1||φ−φ0|

≤ |ζh|(ζ4h+|σ2h−1|2(φ−φ0)2)

=|ζh|5+oP(1)(φ−φ0)2

by Lemma 4.1 and the inequality ab≤a2+b2. Finally, for l= 4, we have OP(n1/2)

( 2 X

h=1

ζ4h|φ−φ0| )

=OP(n1/2) ( 2

X

h=1

{|ζh|5+ (φ−φ0)2} )

since

h|4|φ−φ| ≤ζ8h + (φ−φ0)2

=oP(1)|ζh|5+ (φ−φ)2. If i≥2, we get

OP(n1/2)





2

X

h=1

X

l+s+i=5 l,s,i≥0

h|l2h−1|s|φ−φ0|i





=OP(n1/2)(φ−φ0)2 (4.6.6)

which is due to Lemma 4.1.

Therefore the remainder term (1)n can be simplified to (1)n =OP(n1/2)

2

X

h=1

{|ζh|5 +|ζh|32h−1|+|σ2h−1|3+ (φ−φ0)2}. (4.6.7)

By Lemma 4.1 we can incorporate the terms ml,s,i withl+ 2s+ 4i≥5into the remainder term, e.g.

OP(n1/2)m4,1,0 =OP(n1/24h2h−1) =OP(n1/23h2h−1|, OP(n1/2)m0,4,0 =OP(n1/2)(σ2h−1)4 =OP(n1/2)|σ2h−1|3, OP(n1/2)m2,2,0 =OP(n1/22h2h−1)2 =OP(n1/2)|ζh|(σ2h−1)2

(4.6.8)

and

OP(n1/2)mi,j,2 =OP(n1/2)|ζh|i2h−1|j(φ−φ0)2 =OP(n1/2)(φ−φ0)2 (4.6.9) for i, j ≥0. Adding

OP(n1/2)

2

X

h=1

{|ζh||φ−φ0|+|σ2h−1||φ−φ0|}

to the remainder assures that the terms ml,s,1 with l+ 2s ≥ 1 can be subsumed into the remainder term.

Altogether, we get δt =

4

X

l+2s+4i=1

1 l!s!i!

ζlσs2iφg(Xt|Xt−1; 0, φ0,1)

g(Xt|Xt−1; 0, φ0,1) +tn (4.6.10) with remainder tn fulfilling

n =

n

X

t=1

tn =OP(n1/2)

2

X

h=1

{|ζh|5+|ζh|32h−1|+|σ2h −1|3

+ (φ−φ0)2 +|ζh||φ−φ0|+|σ2h−1||φ−φ0|}.

(4.6.11)

Using the inequality ab≤a2+b2, a, b∈R, we see that

h|32h−1|=|ζh2h2h−1| ≤ |ζh|{ζ4h+ (σ2h−1)2}=|ζh|5+|ζh|(σ2h−1)2.

4.6 Proofs 111

Therefore the remainder term can be written as n =

n

X

t=1

tn =OP(n1/2)

2

X

h=1

{|ζh|5+|ζh|(σ2h−1)2+|σ2h−1|3+ (φ−φ0)2 +|ζh||φ−φ0|+|σ2h−1||φ−φ0|}.

(4.6.12)

With Yt, Zt, Ut, Vt and W1t defined as in the previous chapter we get

δt =t1Yt+t2Zt+t3Ut+t4Vt+t5W1t+tn (4.6.13) with Pn

t=1tn satisfying (4.6.12) and t1 =m1,0,0,

t2 =m2,0,0+m0,1,0, t3 =m3,0,0+ 3m1,1,0,

t4 =m4,0,0+ 6m2,1,0+ 3m0,2,0, t5 =m0,0,1 =φ−φ0.

(4.6.14)

Putting δt into (4.6.3) and noting that the remainders from the square and cubic terms on the right-hand side of the following equation are of the same or higher order than the remainder n from the linear sum, we get

r1n ≤2{t1

n

X

t=1

Yt+t2

n

X

t=1

Zt+t3

n

X

t=1

Ut+t4

n

X

t=1

Vt+t5

n

X

t=1

W1t}

− {t1

n

X

t=1

Yt+t2

n

X

t=1

Zt+t3

n

X

t=1

Ut+t4

n

X

t=1

Vt+t5

n

X

t=1

W1t}2 +2

3{t1

n

X

t=1

Yt+t2

n

X

t=1

Zt+t3

n

X

t=1

Ut+t4

n

X

t=1

Vt+t5

n

X

t=1

W1t}3 +OP(n).

(4.6.15)

As in Chapter 3, one shows that (

t1

n

X

t=1

Yt+t2

n

X

t=1

Zt+t3

n

X

t=1

Ut+t4

n

X

t=1

Vt+t5

n

X

t=1

W1t )3

=oP(n) ( 5

X

l=1

t2l )

. Due to the non-degeneracy of the covariance matrix (Yt, Zt, Ut, Vt, W1t) this implies that the cubic term is dominated by the quadratic term, and the right-hand side of (4.6.15)

reduces to 2{t1

n

X

t=1

Yt+t2

n

X

t=1

Zt+t3

n

X

t=1

Ut+t4

n

X

t=1

Vt+t5

n

X

t=1

W1t}

− {t1 n

X

t=1

Yt+t2 n

X

t=1

Zt+t3 n

X

t=1

Ut+t4 n

X

t=1

Vt+t5 n

X

t=1

W1t}2{1 +oP(1)}

+OP(n).

(4.6.16)

In a next step, we show that

n=oP(1) +oP(n)

5

X

l=1

t2l , (4.6.17)

which is a consequence of the following lemma.

Lemma 4.3. Under the conditions of Lemma 4.1 and the null model we have for h = 1,2, ζ5h =oP

5

X

l=1

|tl|

, ζh2h−1)2 =oP

5

X

l=1

|tl|

and (σ2h−1)3 =oP

5

X

l=1

|tl| .

Proof. Note that by Lemma 4.1tl =oP(1)forl = 1, . . . ,5. Letβh2h+(σ2h−1),h = 1,2, i.e. t2 = (1−α)β1+αβ2. Due to symmetry, we confine our attention toh= 1: The results to be shown forh = 1 hold also true for h= 2.

By the definition of t1 we have

ζ2 ={t1−(1−α)ζ1}/α (4.6.18) and

β2 ={t2−(1−α)β1}/α (4.6.19)

by the definition oft2. Putting (4.6.18) and (4.6.19) into the definition oft3 and replacing σ2h−1by βh−ζ2h therein, we get

t3 = 31−α α

ζ1β1− 2(2α−1) 3α ζ31

+ t1 1 α2

n−2t21+ 3β1α2−3β1α+ 6t1ζ1(1−α)o + t1 6

α2

n−ζ21(1−α)2o + t23

α

t11(α−1) .

4.6 Proofs 113

By Lemma 4.1 we obtain t3 = 31−α

α

ζ1β1− 2(2α−1) 3α ζ31

+oP(|t1|) +oP(|t2|). (4.6.20) Plugging (4.6.18) and (4.6.19) into the definiton of t4 and replacing σ2h −1 by βh −ζ2h therein, it follows that

t4 =31−α α

β21− 2(1−3α+ 3α2) 3α2 ζ41

+ t1 α3

n−t31+ 8(1−α)t21ζ1−12(1−α)2t1ζ21+ 8ζ31(1−α)3o + 3t2

2α−1

α β1+ t2 α

. By Lemma 4.1 we have

t4 = 31−α α

β21− 2(1−3α+ 3α2) 3α2 ζ41

+oP(|t1|) +oP(|t2|). (4.6.21) Multiplying (4.6.20) by β1 +2(2α−1) ζ21 and (4.6.21) by ζ1 and subtracting, we obtain

β1+ 2(2α−1) 3α ζ21

t3−ζ1t4 = 2(1−α)(1−α+α2)

3 ζ51+oP(|t1|) +oP(|t2|).

Note here that the coefficient of t3 in the above equation is oP(1) since we assume α ∈ [δ0,1−δ0] and by Lemma 4.1 ζ21 and β1 are oP(1). The coefficient oft4 is oP(1) since ζ1 is oP(1) (due to Lemma 4.1). Since the coefficients of t3 and t4 are oP(1) and the coefficient of ζ51 is bounded away from 0 (since 1−α+α2 > 0 and 1−α ≥ δ0 which is due to the assumption α∈[δ0,1−δ0]), we get

ζ51 =oP

5

X

l=1

|tl|

. (4.6.22)

Multiplying (4.6.21) by ζ1 and using ζ51 =oP P5 l=1|tl|

, we get ζ1β21 =oP P5 l=1|tl|

since α∈[δ0,1−δ0]. Note that

121−1)2| ≤2|ζ1|(β2141) = 2|ζ1β21|+ 2|ζ51|.

by the definition of β1 and by the inequality (a−b)2 ≤ 2a2 + 2b2, a, b ≥ 0. By ζ1β21 =

oP P5 l=1|tl|

and ζ51 =oP P5 l=1|tl|

we get ζ121−1)2 =oP

5

X

l=1

|tl|

. (4.6.23)

Multiplying (4.6.20) by 2(1−3α+3α2 2)ζ31 and (4.6.21) by β1 and adding yields 2(1−3α+ 3α2)

2 ζ31t31t4 = 31−α α

β31+4(1−3α+ 3α2)(2α−1)

3 ζ61 +oP(|t1|)+oP(|t2|).

Since ζ61 =oP(1)ζ51 =oP P5 l=1|tl|

by Lemma 4.1 and (4.6.22) this leads to 2(1−3α+ 3α2)

2 ζ31t31t4 = 31−α

α β31+oP

5

X

l=1

|tl| .

Given that (1−3α + 3α2)/(3α2) and (1−α)/α are bounded (by the assumption α ∈ [δ0,1−δ0]),

1|3 =oP

5

X

l=1

|tl|

. (4.6.24)

Sinceβ121+ (σ21−1)and using the inequality(a+b)3 ≤4(a3+b3)for a, b≥0we have

21−1|3 =|β121|3 ≤4|β1|3+ 4ζ61 =oP

5

X

l=1

|tl|

, (4.6.25)

where the last equality is due to (4.6.22) and (4.6.24).

Lemma 4.1 gives (φ −φ0)2 = oP(|φ −φ0|) = oP(|t1|) = oP P5 l=1|tl|

and by the same reasoning |ζh||φ −φ0| = oP P5

l=1|tl|

and |σ2h −1||φ −φ0| = oP P5 l=1|tl|

, h = 1,2.

Therefore, we have by Lemma 4.3

n=n1/2oP(

5

X

l=1

|tl|).

Using the inequalities |x| ≤1 +x2 and (a+b)2 ≤2(a2+b2), a, b≥0 repeatedly, we get n =oP(1) +noP(

5

X

l=1

t2l)

as claimed. By the non-degeneracy of the covariance matrix (Yt, Zt, Ut, Vt, W1t)this means

4.6 Proofs 115

that a stochastic upper bound for r1n (which strengthens (4.6.16)) is given by 2{t1

n

X

t=1

Yt+t2

n

X

t=1

Zt+t3

n

X

t=1

Ut+t4

n

X

t=1

Vt+t5

n

X

t=1

W1t}

− {t1

n

X

t=1

Yt+t2

n

X

t=1

Zt+t3

n

X

t=1

Ut+t4

n

X

t=1

Vt+t5

n

X

t=1

W1t}2{1 +oP(1)}

+oP(1).

(4.6.26)

Note thatYt, Zt, Ut, Vt and W1t are mutually orthogonal (see Section 3.6.1). Therefore r1n ≤2t1

n

X

t=1

Yt−t21

n

X

t=1

Yt2{1 +op(1)}

+ 2t2 n

X

t=1

Zt−t22

n

X

t=1

Zt2{1 +op(1)}

+ 2t3

n

X

t=1

Ut−t23

n

X

t=1

Ut2{1 +op(1)}

+ 2t4

n

X

t=1

Vt−t24

n

X

t=1

Vt2{1 +op(1)}

+ 2t5

n

X

t=1

W1t−t25

n

X

t=1

W1t2{1 +op(1)}+oP(1).

(4.6.27)

Using the properties of quadratic functions we have r1n ≤ (Pn

t=1Yt)2 Pn

t=1Yt2 +(Pn t=1Zt)2 Pn

t=1Zt2 + (Pn t=1Ut)2 Pn

t=1Ut2 + (Pn t=1Vt)2 Pn

t=1Vt2 + (Pn

t=1W1t)2 Pn

t=1W1t2 +oP(1).

Proof of Theorem 4.1. Since

2{pln(0.5,ζb0,ζb0,φb0,σb0,σb0)−pln(0.5,0,0, φ0,1,1)}

= 2{ln(0.5,ζb0,ζb0,φb0,bσ0,bσ0)−ln(0.5,0,0, φ0,1,1)}

=(Pn t=1Yt)2 Pn

t=1Yt2 +(Pn t=1Zt)2 Pn

t=1Zt2 +(Pn

t=1W1t)2 Pn

t=1W1t2 +oP(1),

(4.6.28)

we get

2{pln(α, ζ1, ζ2, φ, σ1, σ2)−pln(0.5,ζb0,ζb0,φb0,bσ0,bσ0)}

≤ (Pn t=1Ut)2 Pn

t=1Ut2 +(Pn t=1Vt)2 Pn

t=1Vt2 +oP(1). (4.6.29)

showing that the χ22 distribution serves as a stochastic upper bound for our test statistic EMn(K). It remains to show that this upper bound is also attained, asymptotically. Due to the EM-property it suffices to find adequate parameters α, ζ1, ζ2, σ1, σ2 and φ such that (4.6.29) becomes an equality. This is equivalent to finding a set of values such that tj = ˆtj +oP(n−1/2), j = 1, . . . ,5, where the tj’s are defined in (4.6.14) and

ˆt1 = Pn

t=1Yt Pn

t=1Yt2, ˆt2 = Pn

t=1Zt Pn

t=1Zt2, tˆ3 = Pn

t=1Ut Pn

t=1Ut2, ˆt4 = Pn

t=1Vt Pn

t=1Vt2,tˆ5 = Pn

t=1W1t Pn

t=1W1t2.

Fixing α = 0.5 and ignoring terms of order op(n−1/2) we are searching for parameters ζ1, ζ2, σ1, σ2 and φ satisfying

ˆt1 = 1

2(ζ12), ˆt2 = 1

2(β12), ˆt3 = 3ζ1β1, ˆt4 = 3β12−2ζ14, ˆt5 = (φ−φ0),

(4.6.30)

where tˆ3 = 3ζ1β1+oP(n−1/2) is due to (4.6.20) while ˆt4 = 3β12−2ζ14 +oP(n−1/2) follows from (4.6.21). Define

g(x) = 6x6+ 3ˆt4x2 −tˆ23, x∈R. The third and fourth equation in (4.6.30) imply

g(ζ1) = 6ζ16+ 3ˆt4ζ12−ˆt23

= 6ζ16+ 3(3β12−2ζ1412−(3ζ1β1)2

= 6ζ16+ 9β12ζ12−6ζ16 −9ζ12β12

= 0,

(4.6.31)

Therefore, we have to find a root of g(·) to obtain an adequate ζ˜1. Since g(0) < 0 and g(x) > 0 for x → ∞, there exists a positive root of g(·). Therefore we can choose ζ˜1 to be the smallest positive root of g(·). By (4.6.30) we have β˜1 = ˆt3/(3˜ζ1), ζ˜2 = 2ˆt1 −ζ˜1, β˜2 = 2ˆt2−β˜1 and φ˜= ˆt5−φ0.

By the same reasoning as in Section 3.6, we get ˜tj = ˆtj +oP(n−1/2) and ˆtj = OP(n−1/2), j = 1, . . . ,5. Putting ˜tj, j = 1, . . .5, into the right-hand side of (4.6.27) we see that the

4.6 Proofs 117

upper bound of r1n(·) is also attained. Exemplary, we show this forj = 1:

2˜t1

n

X

t=1

Yt−t˜21

n

X

t=1

Yt2{1 +oP(1)}

= 2 ˆt1+oP(n−1/2)

n

X

t=1

Yt− tˆ1+oP(n−1/2)2 n

X

t=1

Yt2{1 +oP(1)}

= 2(Pn t=1Yt)2 Pn

t=1Yt2 − Pn

t=1Yt

Pn t=1Yt2

2

+OP(n−1/2)oP(n−1/2)

| {z }

=oP(n−1)

+oP(n−1)

n

X

t=1

Yt2{1 +oP(1)}

+op(1)

= 2(Pn t=1Yt)2 Pn

t=1Yt2

(Pn t=1Yt)2 Pn

t=1Yt2 +oP(1)

{1 +oP(1)}+op(1)

= (Pn t=1Yt)2 Pn

t=1Yt2 +oP(1).

Thereby we used the definition of ˆt1, ˆt1 = OP(n−1/2) and the CLT to obtain the second equality. The same argumentation holds true for j = 2,3,4. To obtain the third equality, we used the SLLN. Forj = 5we have to use the CLT for stationary and ergodic martingale differences and the ergodic theorem. Therefore, we have

r1n(0.5,ζ˜1,ζ˜2,φ,˜ σ˜1,˜σ2)

=(Pn t=1Yt)2 Pn

t=1Yt2 +(Pn t=1Zt)2 Pn

t=1Zt2 + (Pn t=1Ut)2 Pn

t=1Ut2 + (Pn t=1Vt)2 Pn

t=1Vt2 + (Pn

t=1W1t)2 Pn

t=1W1t2 +oP(1) and hence

Mn(0)(0.5) = 2{pln(0.5,ζ˜1,ζ˜2,φ,˜ σ˜1,σ˜2)−pln(0.5,ζb0,ζb0,φb0,bσ0,bσ0)}

=r1n(0.5,ζ˜1,ζ˜2,φ,˜ ˜σ1,σ˜2) +r2n

=(Pn t=1Ut)2 Pn

t=1Ut2 + (Pn t=1Vt)2 Pn

t=1Vt2 +oP(1).

By the EM-property, we have

EMn(K) ≥Mn(0)(0.5) = 2{pln(0.5,ζ˜1,ζ˜2,φ,˜ σ˜1,σ˜2)−pln(0.5,ζb0,ζb0,φb0,bσ0,bσ0)}

= (Pn t=1Ut)2 Pn

t=1Ut2 + (Pn t=1Vt)2 Pn

t=1Vt2 +oP(1),

and therefore the limiting distribution of EMn(K) is given by the χ22 distribution.

## 5 Outlook

In this thesis we were mainly concerned with the basic methodological issue of testing for regime switching in Markov-switching autoregressive models. To this end, we extended the work of Cho and White (2007) using penalized likelihood based tests. We gave the limiting distributions of several tests, e.g. the modified (quasi) likelihood ratio test (MQLRT) for testing the hypothesis of no regime switch in regime-switching ARCH models. Since the GARCH(1,1) model is a benchmark model for modeling log returns of asset prices it would be desirable to extend this test to a wider class of models, the so called regime-switching GARCH models, which capture the nice properties of GARCH models and switching au-toregressive models

Xttt; σt2St+

p

X

i=1

αiXt−i2 +

q

X

j=1

βjσt−j2 .

This model does not match our model specification (1.3.3), since σt2 depends on the whole regime path rather than on St, only. This causes a likelihood intractable quickly as soon as the number of observations increases. Following Gray (1996) and Xie and Yu (2005) we consider a so called reduced regime-switching GARCH model

Xttt; σt2St +

p

X

i=1

αiXt−i2 +E

" q X

j=1

βjσt−j2

St−1

# ,

where St−1 is the σ-algebra generated by St−1, St−2, . . . which overcomes the problem of path dependence. While theoretical results such as consistency or asymptotic normality for the MLE in this model were obtained by Xie and Yu (2005), simulations indicate that there should be reasonable hope that the MQLRT for testing the hypothesis of no regime switch in e.g. a reduced switching GARCH(1,1) asymptotically admits a mixture of a point mass at zero and aχ21 distribution with equal weights.

Rejecting the hypothesis of one regime we know that there will be at least two regimes.

In this case, the determination of the true number of regimes remains an open problem.

It would be desirable to find an adequate alternative to the time demanding Bootstrap approach introduced by McLachlan (1987) in the case of mixture models. In a first step, testing for two states in a Markov-switching autoregressive model against (at least) three states could be performed via a MQLRT extending the work of Chen, Chen and Kalbfleisch (2004) for mixture models and of Dannemann and Holzmann (2008) for HMMs, especially in case of model (2.2.3).

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## Zusammenfassung

Eine Vielzahl von Zeitreihenmodellen, wie z.B. lineare autoregressive oder ARCH-Modelle, wird verwendet, um das Verhalten von ökonomischen und Finanzzeitreihen zu analysieren.

Da sich jedoch das Verhalten von Zeitreihen über die Zeit häufig ändert, beschreiben solche Zeitreihenmodelle die Daten möglicherweise nicht zufriedenstellend.

Das Markov-Switching-Modell von Hamilton (1989) ist eines der bekanntesten Regime-Switching-Modelle. Dieses Modell bildet mehrere Strukturen ab, die das Verhalten der Zeitreihe in verschiedenen Zuständen charakterisieren. Während der switchende Mecha-nismus im ursprünglichen Markov-Switching-Modell in ein lineares autoregressives Mod-ell aufgenommen wurde, untersuchten Cai (1994) und Hamilton und Susmel (1994) ver-schiedene ARCH-Modelle mit Markov-Switching. In Markov-Switching-Modellen wird der switchende Mechansimus durch eine latente Variable, die einer Markov-Kette der Ord-nung 1 folgt, gesteuert. Deshalb ist u.a. die Bestimmung der Anzahl der Zustände der versteckten Markov-Kette von großer Bedeutung. In dieser Arbeit beschäftigen wir uns insbesondere mit der grundlegenden methodischen Fragestellung des Testens auf Regime-Switching in diversen Markov-Regime-Switching autoregressiven Modellen. Da unter der Hy-pothese Parameter des vollen Modells nicht identifizierbar sind, ist die asymptotische Verteilung des entsprechenden Likelihood-Quotienten-Tests nicht standard. Dieses Prob-lem tritt auch schon in der eng verwandten Fragestellung des Testens auf Homogenität in Zwei-Komponenten-Mischungsmodellen auf. Um das Problem der Nicht-Identifizierbarkeit zu lösen, entwickelten Chen, Chen und Kalbfleisch (2001) einen penalisierten Likelihood-Quotienten-Test und zeigten, dass die entsprechende Teststatistik eine einfache asymp-totische Verteilung hat. Zusätzliche Schwierigkeiten treten auf, wenn wir die Markov-Abhängigkeitsstruktur in die Teststatistik aufnehmen. Deshalb schlugen Cho und White (2007) einen quasi Likelihood-Quotienten-Test auf Regime Switching in autoregressiven Modellen vor, der die Abhängigkeitsstruktur in der versteckten Markov-Kette unter der Alternative vernachlässigt. Wir erweitern diese Vorgehensweise, indem wir penalisierte Likelihood-basierte Tests entwickeln, um Tests mit einfachen asymptotischen Verteilungen zu erhalten.

In Kapitel 1 stellen wir Markov-Switching autoregressive und eng verwandte Modelle vor und diskutieren die Methodik, die wir im Folgenden benutzen.

Der modifizierte Likelihood-Quotienten-Test, der von Chen, Chen und Kalbfleisch (2001) eingeführt wurde, ist eine etablierte Methode zum Testen auf Homogenität in endlichen Mischungsmodellen. In Kapitel 2 erweitern wir diesen Test auf Markov-Switching autore-gressive Modelle mit univariatem switchenden Parameter, die gewisse

Regularitätsbeding-ungen erfüllen. Diese BedingRegularitätsbeding-ungen sind z.B. für lineare switching autoregressive Mod-elle mit switchendem Skalenparameter und normalverteilten Innovationen erfüllt. Wir zeigen, dass die asymptotische Verteilung des modifizierten quasi Likelihood-Quotienten-Tests unter der Hypothese eine Mischung aus einer Punktmasse in0und einerχ21-Verteilung ist. Schließlich führen wir einen verwandten Test, den sog. EM-Test, ein, der die gleiche asymptotische Verteilung wie der modifizierte quasi Likelihood-Quotienten-Test aufweist.

Für Anwendungen ist das lineare switching autoregressive Modell mit switchendem In-tercept und normalverteilten Innovationen von großer Bedeutung, siehe Hamilton (2008).

In Kapitel 3 geben wir einen Test auf Homogenität in diesem Modell an. Allerdings ist das Studium asymptotischer Eigenschaften von Teststatistiken, die auf der (penalisierten) Likelihood basieren, sehr schwierig, da σ2f(x;µ,σ)2µ = ∂f(x;µ,σ)∂σ für die Normalverteilung gilt.

Hierbei ist f(x;µ, σ) die Dichte der Normalverteilung mit Erwartungswert µ und Stan-dardabweichung σ > 0. Dieses Problem tritt bereits beim Testen auf Homogenität in ho-moskedastischen Mischungen univariater Normalverteilungen auf. Chen und Li (2009) ent-wickelten hierfür einen Test. In Kapitel 3 übertragen wir diesen Test auf lineare switching autoregressive Modelle mit normalverteilten Innovationen, in denen der Intercept gemäß des zugrundeliegenden Regimes switcht. Wir zeigen, dass die asymptotische Verteilung der entsprechenden Teststatistik unter der Hypothese eine einfache Funktion der verschobenen χ21- und 12χ20+12χ21-Verteilung ist. Ferner schlagen wir einen Test gegen feste Gewichte unter der Alternative vor und berechnen die asymptotische Verteilung dieser Teststatistik unter der Hypothese. Wir wenden die in Kapitel 2 und 3 entwickelten Methoden auf die saisonal bereinigten Quartalsdaten des U.S. BIPs von 1947(1) bis 2002(3) an und finden einen Regime Switch in der Volatilität der Wachstumsrate. Schließlich teilen wir die Zeitreihe in zwei Teilzeitreihen 1947(1)-1984(1) und 1984(2)-2002(3) auf und finden keine klare Evidenz für einen Regime Switch im Intercept in einem linearen autoregressiven Modell in diesen Teilreihen.

In Kapitel 4 beschäftigen wir uns mit Tests auf Homogenität in einem linearen switching autoregressiven Modell, in dem sowohl Intercept als auch Varianz der normalverteilten Innnovationen switchen dürfen. Wir erweitern den sog. EM-Test von Chen und Li (2009) zum Testen auf Homogenität in einem Normalverteilungsmischungsmodell mit verschiede-nen Erwartungswerten und Varianzen unter der Alternative. Wir zeigen, dass die asymp-totische Verteilung der Teststatistik unter der Hypothese eine χ22-Verteilung ist. Da der EM-Test die gleiche asymptotische Verteilung aufweist, wenn wir unter der Alternative das Gewicht α = 1/2 festhalten, schlagen wir auch einen Test gegen festes Gewicht α = 1/2 unter der Alternative vor. Wir wenden unsere Methoden auf die monatlichen log-Returns der IBM-Aktie an und finden Evidenz für 2 Zustände: Regime 1 mit niedrigerem Intercept und höherer Varianz und Regime 2 mit höherem Intercept und niedriger Varianz.

Outline

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